Minimum Substring Partition of Equal Character Frequency Python/Java solution

Implement a dynamic programming array where dp[i] stores the minimum partitions for the prefix ending at i. For each starting index, maintain a 26-length frequency array while expanding the substring. When the substring satisfies maxFreq * distinct == length, update the DP state. The same logic works in Python, Java, C++, and other languages.

What is the best approach for Minimum Substring Partition of Equal Character Frequency?

The most common approach uses dynamic programming with character frequency counting. For every starting index, expand the substring and maintain counts of characters. A substring is valid when the length equals maxFrequency multiplied by the number of distinct characters. This leads to an O(n^2) solution with O(n) DP space.

Is Minimum Substring Partition of Equal Character Frequency asked at Google/Amazon/Meta?

Problems involving substring partitioning, frequency balancing, and dynamic programming frequently appear in interviews at companies like Google, Amazon, and Meta. Variants that combine string processing with DP are especially common in mid‑level and senior coding interviews.

What data structure is used in Minimum Substring Partition of Equal Character Frequency?

The solution mainly uses a frequency array or hash table to count characters, along with a dynamic programming array to store the minimum number of partitions. These structures allow constant-time updates while scanning substrings.

What is the time complexity of Minimum Substring Partition of Equal Character Frequency?

The typical solution runs in O(n^2) time. For each starting index you extend the substring while maintaining character frequencies, which takes constant time per extension because the alphabet size is fixed. Space complexity is O(n) for the dynamic programming array.

Description Hints 2 Solution Video 27:29

3144. Minimum Substring Partition of Equal Character Frequency

Q: Is Minimum Substring Partition of Equal Character Frequency easy or hard?

The problem is rated Medium on LeetCode. The challenge comes from recognizing the equal-frequency property and combining substring validation with dynamic programming to minimize partitions.

Q: How to solve Minimum Substring Partition of Equal Character Frequency in O(n)?

An O(n) solution is not generally feasible because every potential substring boundary may need to be checked. Most accepted solutions rely on O(n^2) dynamic programming combined with frequency counting. The optimization comes from validating substrings in constant time using maintained counts instead of recomputing them.

Medium40.5% acceptance4 companies

Given a string s, you need to partition it into one or more balanced substrings. For example, if s == "ababcc" then ("abab", "c", "c"), ("ab", "abc", "c"), and ("ababcc") are all valid partitions, but ("a", "bab", "cc"), ("aba", "bc", "c"), and ("ab", "abcc") are not. The unbalanced substrings are bolded.

Return the minimum number of substrings that you can partition s into.

Note: A balanced string is a string where each character in the string occurs the same number of times.

Example 1:

Input: s = "fabccddg"

Output: 3

Explanation:

We can partition the string s into 3 substrings in one of the following ways: ("fab, "ccdd", "g"), or ("fabc", "cd", "dg").

Example 2:

Input: s = "abababaccddb"

Output: 2

Explanation:

We can partition the string s into 2 substrings like so: ("abab", "abaccddb").

Constraints:

1 <= s.length <= 1000
s consists only of English lowercase letters.

Asked at 4 companies

Mitsogo Grammarly Morgan Stanley Microsoft

Topics

Hints open one at a time

Hint 1

Let dp[i] be the minimum number of partitions for the prefix ending at index i + 1.

Hint 2

dp[i] can be calculated as the min(dp[j]) over all j such that j < i and word[j+1…i] is valid.

27:29

Still stuck?Read 5 approaches, or watch a walkthrough.

Open solution